A short bar magnet of moment $3.2 A m^{2}$ lies in magnetic meridian with its north pole pointing north. A neutral point is obtained at a distance of $20$ cm from the centre of the magnet. Find the horizontal.

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Solution

At null point, magnet's magnetic field is balanced by the earth's field.
r=20cm=0.2m
$B_{m a g n e t} = B_{H}$

$\frac{μ_{0}}{4 π} \frac{M}{r^{3}} = B_{H}$

$10^{- 7} * \frac{3.2}{(0.2)^{3}} = B_{H}$
$∴ B_{H} = 4 * 10^{- 5} T$

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A bar magnet of magnetic moment $1.2 A m^{2}$ is placed in the magnetic meridian with its south pole pointing the north. If a neutral point is found at a distance of $20 c m$ from the centre of the magnet, the value of the horizontal component of the earth's magnetic field is

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A short bar magnet of moment 3.2Am2 lies in magnetic meridian with its north pole pointing north. A neutral point is obtained at a distance of 20cm from the centre of the magnet. Find the horizontal.

At null point, magnet's magnetic field is balanced by the earth's field. r=20cm=0.2mBmagnet=BH μ04πMr3=BH 10−7∗3.2(0.2)3=BH∴BH=4∗10−5T

A short bar magnet of moment $3.2 A m^{2}$ lies in magnetic meridian with its north pole pointing north. A neutral point is obtained at a distance of $20$ cm from the centre of the magnet. Find the horizontal.

At null point, magnet's magnetic field is balanced by the earth's field.
r=20cm=0.2m
$B_{m a g n e t} = B_{H}$

$\frac{μ_{0}}{4 π} \frac{M}{r^{3}} = B_{H}$

$10^{- 7} * \frac{3.2}{(0.2)^{3}} = B_{H}$
$∴ B_{H} = 4 * 10^{- 5} T$